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w^2+3w-108=0
a = 1; b = 3; c = -108;
Δ = b2-4ac
Δ = 32-4·1·(-108)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{441}=21$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-21}{2*1}=\frac{-24}{2} =-12 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+21}{2*1}=\frac{18}{2} =9 $
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